Difficulty: Medium
Correct Answer: η = 2V / (V + V_j)
Explanation:
Introduction:
Jet propulsion in watercraft relies on momentum exchange: water is ingested and expelled at a different velocity to create thrust. Propulsive efficiency compares useful power (thrust * boat speed) to the rate of kinetic energy imparted to the water stream. Knowing the correct efficiency expression is key to optimizing jet speed for best economy.
Given Data / Assumptions:
Concept / Approach:
For an ideal momentum jet, thrust T equals mass flow rate m_dot times change in axial velocity between intake (≈ V forward) and discharge (≈ −V_j aft), giving T = m_dot * (V + V_j). Useful power is T * V. The rate of kinetic energy imparted to the stream is (1/2) m_dot * [(V + V_j)^2 − V^2] = (1/2) m_dot * (V_j^2 + 2 V V_j). The well-known simplification yields η = 2V / (V + V_j).
Step-by-Step Solution:
1) Axial velocity change: Δu = V + V_j.2) Thrust: T = m_dot * Δu.3) Useful power: P_use = T * V = m_dot * (V + V_j) * V.4) Jet power increase (ideal): P_jet = (1/2) m_dot * [(V + V_j)^2 − V^2].5) Efficiency: η = P_use / P_jet = 2V / (V + V_j).
Verification / Alternative check:
Check limits: if V_j → V (small slip), η → 1; if V_j » V, η drops, matching intuition that excessively fast jets waste energy.
Why Other Options Are Wrong:
V/V_j and (V_j − V)/V_j ignore momentum-power balance.
2V_j/(V + V_j) inverts the dependence.
(V + V_j)/(2V) is the reciprocal of the correct form.
Common Pitfalls:
Mixing absolute and relative velocities; omitting the inflow kinetic energy carried on board with the incoming stream.
Final Answer:
η = 2V / (V + V_j)
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